Visual walkthrough — Thermal design power (TDP)

6.4.3 · D2 · Hardware › Power, Thermal & Reliability › Thermal design power (TDP)

Tumne shayad Thermal design power (TDP) parent note pe headline formula dekha hoga:

$$\text{TDP}=\frac{T_{\text{j}}-T_{\text{ambient}}}{R_{\theta ,\text{total}}}$$

Lekin yeh aata kahan se hai? Kyun ek temperature difference ko resistance se divide karne par heat ka flow milta hai? Is page pe hum yeh equation bilkul scratch se banate hain — koi pehle se formula assume nahi — har idea ke liye ek picture use karke. Akhir mein tum poori cooling law ko ek napkin pe draw kar paoge.

Pehle us pehle word pe agree karte hain jo hum use karenge.

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Step 1 — Ek chip heat ka ek leaky bucket hai

KYA HAI. CPU die (silicon ka chota sa square) ko ek tiny stove samjho. Har second yeh ek fixed amount of heat produce karta hai — us number ko $P$ watts kaho. Woh heat kahi na kahi jaani chahiye, warna yeh pile up ho jaati hai.

KYUN. Cooling ki baat karne se pehle, hume bilkul clear hona chahiye ki chip ek constant heat source hai. Ek chip jitni electrical power draw karti hai uska almost sabhi hissa heat mein badal jaata hai (yeh koi lifting, koi motion nahi karta) — isliye jab cheezein settle ho jaati hain toh "power in" equals "heat out" hota hai. Yeh ek simple fact hi baad mein hume "TDP" (ek heat number) ko electrical-power reasoning se likhne deta hai.

PICTURE. Laal die heat upar ki taraf pour kar rahi hai. Agar kuch usse le nahi jaata, temperature hamesha badhti rahegi.

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Step 2 — Temperature ek "height" hai, aur heat downhill flow karti hai

KYA HAI. Har point ko ek temperature do: garma chip die $T_{\text{j}}$ pe baithta hai ("$\text{j}$" matlab junction, silicon ka kaam karne wala andar ka hissa), aur door ka room air $T_{\text{ambient}}$ pe baithta hai. Heat hamesha bade number se chhote number ki taraf slide karti hai.

KYUN. Heat flow ko move karne ka ek reason chahiye, bilkul waise jaise paani ko downhill slope chahiye. Woh reason hai temperature difference:

$$\Delta T=\underset{\text{hot chip}}{\underbrace{T_{\text{j}}}}-\underset{\text{cool room air}}{\underbrace{T_{\text{ambient}}}}$$

• $\Delta T$ (padho "delta-T") — chip se air tak temperature ki drop. Yeh woh push hai jo heat ko bahar dhakelta hai.
• $T_{\text{j}}$ — pahaad ki choti.
• $T_{\text{ambient}}$ — pahaad ka neeche.

Hum yahan subtraction choose karte hain (ratio nahi, sum nahi) kyunki heat ko jo move karta hai woh purely hai chip room se kitna zyada garam hai — heights ka ek difference.

PICTURE. $T_{\text{j}}$ se $T_{\text{ambient}}$ tak ka step jitna bada, heat utni hi tezi se bahar push hoti hai.

Recall

Woh kaunsi ek quantity hai jo chip se heat bahar nikalne ka "push" hai? ::: Temperature difference $\Delta T=T_{\text{j}}-T_{\text{ambient}}$.

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Step 3 — Kuch flow ko resist karta hai: thermal resistance

KYA HAI. Heat teleport nahi hoti bahar. Usse silicon se, metal case se, paste se, heatsink se, aur air mein ooze karna padta hai. Har layer flow ke saath larti hai. Hum us saari ladaai ko ek number $R_{\theta }$ mein bundle karte hain, thermal resistance, jise °C per watt (°C/W) mein measure karte hain.

KYUN. Resistance invent kyon karo? Kyunki experiment ek beautifully simple pattern dikhata hai: usi material se zyada heat push karne ke liye, tumhe proportionally bada temperature drop chahiye. Woh "ek watt flow per kitne degrees ka drop" exactly hai jo $R_{\theta }$ count karta hai.

$$R_{\theta }=\frac{\Delta T}{P}⟺\underset{\text{degrees}}{\underbrace{\Delta T}}=\underset{\text{watts of heat}}{\underbrace{P}}\times \underset{\text{°C per watt}}{\underbrace{R_{\theta }}}$$

Units padho feel karne ke liye: $\text{watts}\times \frac{\text{°C}}{\text{watt}}=\text{°C}$. Watts cancel ho jaate hain, degrees bach jaate hain — arithmetic literally ek temperature bana deta hai.

• Chhota $R_{\theta }$ = aasaan raasta = heat sirf ek tiny temperature step se nikal jaati hai. Achha cooler.
• Bada $R_{\theta }$ = bandh raasta = usi heat ko bahar force karne ke liye tumhe ek huge (dangerous) temperature step chahiye. Bura cooler.

PICTURE. Do pipes same paani ka flow carry karti hain: moti pipe (low resistance) ko barely koi slope chahiye; patli pipe (high resistance) ko steep slope chahiye.

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Step 4 — Balance: heat in must equal heat out

KYA HAI. Chip $P$ watts andar pour karta hai. Path $\Delta T/R_{\theta }$ watts bahar drain karta hai. Ek moment ruko aur yeh dono ek tie mein settle ho jaate hain — ek steady state — jahan temperature change karna band kar deta hai.

KYUN. Agar chip path se zyada heat banata, temperature badhti, jo $\Delta T$ raise karta, jo drain increase karta, jab tak woh match nahi kar lete. Agar path bahut tezi se drain karta, temperature girta, $\Delta T$ shrink karta, drain slow karta, jab tak woh match nahi kar lete. System apne aap us point pe correct ho jaata hai:

$$\underset{\text{heat produced/sec}}{\underbrace{P}}=\underset{\text{heat drained/sec}}{\underbrace{\frac{\Delta T}{R_{\theta }}}}$$

Yeh equilibrium hi woh reason hai ki chip ek stable temperature tak pahunchti hai instead of explode karne ke.

PICTURE. Bucket mein ab ek inflow ($P$) bhi hai aur ek drain bhi jiska speed paani ki height ke saath badhta hai. Level wahan park ho jaata hai jahan inflow = outflow.

Recall

Steady state mein, chip ke generated heat ko kya balance karta hai? ::: Path se drained heat, $\Delta T/R_{\theta }$ — woh equal hain, isliye temperature badhna band ho jaata hai.

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Step 5 — Path mein kai resistances ek row mein hain

KYA HAI. Escape route ek wall nahi balki teen series mein stack hain: junction→case, case→sink, sink→air. Series resistances add ho jaate hain.

KYUN. Heat ko teeno ek ke baad ek cross karne padte hain — jaise ek hallway mein teen doorways se guzarna. Total delay har doorway ki delay ka sum hai. Kuch skip nahi hota, isliye sum se kuch drop nahi hota:

$$R_{\theta ,\text{total}}=\underset{\text{silicon}\to \text{metal case}}{\underbrace{R_{\theta ,\text{JC}}}}+\underset{\text{paste layer}}{\underbrace{R_{\theta ,\text{CS}}}}+\underset{\text{heatsink}\to \text{air}}{\underbrace{R_{\theta ,\text{SA}}}}$$

• $R_{\theta ,\text{JC}}$ — Junction-to-Case, chip mein baked in; tum ise change nahi kar sakte.
• $R_{\theta ,\text{CS}}$ — Case-to-Sink, thermal paste. Sasta paste = zyada = worse.
• $R_{\theta ,\text{SA}}$ — Sink-to-Ambient, fan-and-fins wala part jo tum khareedtay ho.

PICTURE. Teen temperature steps ek staircase mein, har step ki height uske apne resistance se set hoti hai jo same heat $P$ carry kar raha hai.

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Step 6 — Balance ko rearrange karke TDP cooling law banao

KYA HAI. Step 4 ka steady-state balance lo, $R_{\theta }$ ko Step 5 ke total se replace karo, aur sustained heat number $P$ ko TDP naam do — woh heat jo manufacturer promise karta hai ki chip ek realistic heavy load mein banayegi.

KYUN. TDP defined hai us sustained heat rate ke roop mein. Ise balance mein plug karna physics ko ek design contract mein badal deta hai: "cooling, tumhe itne watts drain karne chahiye."

$$\boxed{\text{TDP}=\frac{T_{\text{j}}-T_{\text{ambient}}}{R_{\theta ,\text{total}}}}$$

Term by term, yeh sirf Step 4 hai achhe labels ke saath:

• upar $T_{\text{j}}-T_{\text{ambient}}$ = temperature push (Step 2),
• neeche $R_{\theta ,\text{total}}$ = total path resistance (Step 5),
• baayein $\text{TDP}$ = woh heat jo flow karni chahiye (Steps 1 & 4).

PICTURE. Finished flow diagram: upar push, neeche resistance, side se heat stream kar rahi hai.

Ab ise flip karo design form paane ke liye. Hum usually TDP jaante hain (box se), $T_{\text{j,max}}$ (datasheet se, ~90–105 °C), aur room air $T_{\text{ambient}}$. Hum solve karte hain us resistance ke liye jo hum rakh sakte hain:

$$R_{\theta ,\text{total}}\le \frac{T_{\text{j,max}}-T_{\text{ambient}}}{\text{TDP}}$$

Hum ≤ likhte hain (= nahi) kyunki ek lower resistance chip ko limit se thanda chalata hai — hamesha welcome hai. Limit se upar kuch bhi $T_{\text{j}}$ ko $T_{\text{j,max}}$ se aage push karta hai aur Thermal throttling trigger karta hai.

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Step 7 — Edge cases: formula apni breaking points pe

Har law ko test karna chahiye jahan lagta hai ki woh toot sakta hai. Teen degenerate cases:

Case A — Zero resistance ($R_{\theta ,\text{total}}\to 0$). Ek tiny number se divide karne par ek huge allowed TDP milta hai. Sense banta hai: ek perfect, resistance-free cooler kisi bhi amount of heat ko bina kisi temperature rise ke drain kar sakta hai. Yeh unreachable dream hai — tum approach kar sakte ho (liquid nitrogen), touch kabhi nahi.

Case B — Garam room ($T_{\text{ambient}}\to T_{\text{j,max}}$). Top $T_{\text{j,max}}-T_{\text{ambient}}\to 0$, isliye allowed resistance $\to 0$. Agar room pehle se hi chip ki limit jitni garma hai, toh koi push bacha hi nahi — duniya ka koi cooler kaam nahi aayega. Isliye Data center cooling intake air ko thanda rakhne ke liye itni mehnat karta hai: room air ka har ek degree zyada garma hona tumhara $\Delta T$ budget cheen leta hai.

Case C — Room limit se zyada garam. Tab $T_{\text{j,max}}-T_{\text{ambient}}<0$; formula ek negative resistance nikalti hai, jo physically impossible hai. Honest reading: koi cooler kaam nahi karega. Heat apne aap upar nahi beh sakti, isliye chip overheat hone ke liye doomed hai. Math ka "negative" flag karna uska tarika hai chillane ka "give up."

PICTURE. Allowed-resistance budget zero tak shrink ho raha hai aur negative ho raha hai jab room garam hoti hai.

Recall

Kyun ek garam room cooling impossible bana deta hai even with ek giant heatsink? ::: Kyunki cooling temperature difference $\Delta T$ pe chalti hai. Jab ambient $T_{\text{j,max}}$ tak pahunchta hai, $\Delta T\to 0$, isliye heatsink kitna bhi achha ho koi heat push nahi hoti bahar.

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Ek-picture summary

Upar sab kuch ek single "voltage-divider style" picture mein collapse ho jaata hai: ek heat current $\text{TDP}$ teen stacked resistances se flow kar raha hai, temperature ko upar $T_{\text{j}}$ se neeche $T_{\text{ambient}}$ tak drop karta hai, har resistance ke liye ek step.

Upar se neeche padho: garme junction se shuru karo, har resistance ke liye ek temperature step subtract karo (per layer $P\times R$), aur jab tak tum exit naho tab tak tum room pe ya neeche land karne chahiye — warna chip cook ho jaati hai. Chhota heatsink resistance $R_{\theta ,\text{SA}}$ choose karna uske step ko shrink karta hai, zyada headroom chodta hai.

Recall Feynman retelling — kisi dost ko samjhao jaise

Chip ek stove hai jo fixed number of watts of heat bana rahi hai. Heat tabhi move hoti hai jab ek side doosri se zyada garam ho, aur yeh kitna move hoti hai depend karta hai ki escape path kitna "clogged" hai — woh clog thermal resistance hai. Clogs ko (silicon, paste, heatsink) ek row mein string karo aur woh add up ho jaate hain. Balance pe, banai gayi heat drained heat ke equal hoti hai, isliye chip ka temperature sirf room temperature plus (watts × total resistance) hai. Ise ulta karo aur yeh cooler-builders ko bataata hai ki unhe maximum kitna resistance allowed hai. Aur agar room kabhi chip ki limit jitni garam ho jaaye, push khatam ho jaata hai aur koi cooler tumhe bacha nahi sakta — equation zero ho jaake phir negative ho jaake tumhe warn karta hai.

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Yeh bhi dekho: Heatsink design and thermal resistance · Power consumption in CMOS circuits · Thermal throttling · Turbo Boost and power states · Laptop thermal design